Question

In the following exercises, evaluate the rational expression for the given values.

$$\begin{gathered} \frac{z^2+3z-10}{z^2-1} \\ \left(a\right)z=0 \\ \left(b\right)z=2 \\ \left(c\right)z=-2 \end{gathered}$$

Short answer

The required answers are

$$\begin{gathered} \mathrm{Part}\left(\mathrm{a}\right)10 \\ \mathrm{Part}\left(\mathrm{b}\right)0 \\ \mathrm{Part}\left(\mathrm{c}\right)-4 \end{gathered}$$

Step-by-step solution

Part(a) Step 1. Given Information 

The given expression is $\frac{z^2+3z-10}{z^2-1}$

Part (a) Step 2. Calculation   

Substitute 0 for z,

$$\begin{gathered} \frac{z^2+3z-10}{z^2-1}=\frac{{\left(0\right)}^2+3\left(0\right)-10}{{\left(0\right)}^2-1} \\ =\frac{-10}{-1} \\ =10 \end{gathered}$$

Part (b) Step 1. Calculation  

Substitute 2 for z,

$$\begin{gathered} \frac{z^2+3z-10}{z^2-1}=\frac{{\left(2\right)}^2+3\left(2\right)-10}{{\left(2\right)}^2-1} \\ =\frac{4+6-10}{4-1} \\ =\frac{0}{3} \\ =0 \end{gathered}$$

Part (c) Step 1. Calculation    

Substitute $-2$ for z,

$$\begin{gathered} \frac{z^2+3z-10}{z^2-1}=\frac{{(-2)}^2+3(-2)-10}{{(-2)}^2-1} \\ =\frac{4-6-10}{4-1} \\ =\frac{-12}{3} \\ =-4 \end{gathered}$$